`int_(0)^(2) sqrt(2-x)dx`.

Answer

To solve the integral \( \int_{0}^{2} \sqrt{2 - x} \, dx \), we will follow these steps:

### Step 1: Substitution
Let's use the substitution \( u = 2 - x \). Then, we have:
\[ du = -dx \quad \text{or} \quad dx = -du \]
When \( x = 0 \), \( u = 2 \) and when \( x = 2 \), \( u = 0 \). Thus, the limits of integration change from \( 0 \) to \( 2 \) to \( 2 \) to \( 0 \).

### Step 2: Rewrite the Integral
Substituting into the integral, we get:
\[ \int_{0}^{2} \sqrt{2 - x} \, dx = \int_{2}^{0} \sqrt{u} \, (-du) = \int_{0}^{2} \sqrt{u} \, du \]

### Step 3: Evaluate the Integral
Now, we can evaluate the integral:
\[ \int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \]
Now, we apply the limits from \( 0 \) to \( 2 \):
\[ \left[ \frac{2}{3} u^{3/2} \right]_{0}^{2} = \frac{2}{3} (2^{3/2}) - \frac{2}{3} (0^{3/2}) = \frac{2}{3} \cdot 2\sqrt{2} - 0 = \frac{4\sqrt{2}}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{2} \sqrt{2 - x} \, dx \) is:
\[ \frac{4\sqrt{2}}{3} \]

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